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Proof step 10 from proof bitsand-len-false (previous step)

bits = bits2 + boolvar ', # bits1 < # bits, ∀ bits, bits0. # bits0 < # bits1 → # bits0 < # bits → # bitsand(bits, false ' + bits0) = 1 + # bits0 ∨ bits = [] ⊦ # bits = 1, bits1 = [], bits = [], # bitsand(bits2, false ' + bits1.butlast) = # bits1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# bits1 ≤ # bits2, ∀ bits, bits0. # bits0 < # bits1 → # bits0 < # bits → # bitsand(bits, false ' + bits0) = 1 + # bits0 ∨ bits = [] ⊦ bits1 = [], # bitsand(bits2, false ' + bits1.butlast) = # bits1, bits2 = []

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ bits + boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;